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Q.

Number of electrons present in $4 f$ orbital of ${Ho}^{+ 3}$ ion is ______.
(given atomic number of $Ho = 67$ )

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answer is 10.

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Detailed Solution

The atomic number of $Ho$ is known to be 67, so its electronic configuration will be written as $[Xe] 4 f^{11} 6 s^{2}$ .
The ion given in the question is ${Ho}^{+ 3}$ , so it is the ion of $Ho$ with +3 electron charge, that is, its three outermost electrons are removed from their respective orbitals. So the electronic configuration of ${Ho}^{+ 3}$ would be $[Xe] 4 f^{10}$ .

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